Question

A drop of liquid of density $\rho$ is floating half-immersed in a liquid of density $d$. If $\sigma$ is the surface tension, then the diameter of the drop of the liquid is

• A. $\sqrt{\frac{\sigma}{g(2\rho - d)}}$
• B. $\sqrt{\frac{2\sigma}{g(2\rho - d)}}$
• C. $\sqrt{\frac{8\sigma}{g(2\rho - d)}}$
• D. $\sqrt{\frac{12\sigma}{g(2\rho - d)}}$

Hint:

Surface tension and buoyancy forces balance out when a drop floats. The diameter can be calculated by this relation involving surface tension and densities.

Answer 1

The Correct Option is D

For a liquid drop floating in another liquid, the buoyancy force is balanced by the surface tension. The relationship between the diameter of the drop and the surface tension is given by: \[ d = \sqrt{\frac{12 \sigma}{g(2\rho - d)}} \]